Q.Find the third quartile $Q_3$ for the continuous distribution: classes $0\text{–}20, 20\text{–}40, 40\text{–}60, 60\text{–}80, 80\text{–}100$ with frequencies $8, 12, 20, 7, 3$.

Build the $cf$ column ($N = 50$), locate the $Q_3$ class using $\frac{3N}{4}$, then interpolate (note $h = 20$ here).

$Q_3 = 40 + \frac{37.5 - 20}{20}\times 20 = 57.5$.

Build the less-than cumulative frequencies:

$N = 50$, so $\dfrac{3N}{4} = \dfrac{150}{4} = 37.5$. The first $cf \ge 37.5$ is $40$ → the $Q_3$ class is $40\text{–}60$: $L = 40, cf = 20, f = 20, h = 20$.

Independent check. The class width is $h = 20$, and $\tfrac{17.5}{20} = 0.875$ of it is $17.5$; adding to $L = 40$ gives $57.5$, which lies inside $40\text{–}60$ as required ✓. Since $Q_3$ should have $75\%$ ($37.5$ of $50$) below it, and the $cf$ reaches $37.5$ partway through the $40\text{–}60$ class, the value near the upper end of that class is consistent.

Here $h = 20$, not $10$ — a reminder that $h$ is the actual width of the partition class. Had one wrongly used $h = 10$, the fraction $0.875$ would give $8.75$ and a wrong $Q_3 = 48.75$; always read $h$ from the class itself.

Assuming a class width of $10$ out of habit. This distribution has width $20$, and the interpolation must use it. Also, take $cf = 20$ (the class before), not $40$ (the $Q_3$ class's own cumulative total).